Calculate the magnetic moment of K_4[Mn(CN)_6] complex.

The correct answer is Option (3) → 1.73 BM

The spin-only magnetic moment depends on the number of unpaired electrons, n:

    \( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)

To find n, first find the oxidation state of manganese. K is +1 and each \( \ce{CN^-} \) is −1:

    \( \displaystyle 4(+1) + x + 6(-1) = 0 \Rightarrow x = +2 \)

Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Mn^2+} \): \( \mathrm{3d^{5}} \).

\( \ce{CN^-} \) is a strong field ligand. In the octahedral complex the splitting \( \Delta_o \) is larger than the pairing energy, so all five electrons go into the lower \( t_{2g} \) set: \( t_{2g}^{5}\,e_g^{0} \), which is two pairs and one unpaired electron (a low spin, inner orbital \( \mathrm{d^{2}sp^{3}} \) complex).

    \( \displaystyle \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\ \mathrm{BM} \)

So the magnetic moment of \( \ce{K4[Mn(CN)6]} \) is 1.73 BM.